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3 years ago

High pressure system the air moves (up/down) and brings (good/bad) weather

Physics

1 answer:

satela [25.4K]3 years ago

3 0

Answer:

high pressure system in the air moves up and brings good weather

give me thanks >:(

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Please help me-. it’s part of my major grade.

Serga [27]

Answer:

Explanation:

6 0

3 years ago

Read 2 more answers

Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 64.7 N, Jill pulls with 61.5 N in the

USPshnik [31]

Answer:

F = 241.8 N θ = -18.7

Explanation:

To find the net force, we use Newton's second law, for this we decompose the force using trigonomer

Force 2

F2 = 61.5 N

cos 45 = F₂ₓ / F2 F₂ₓ = F2 cos 45

sin 45 = F₂_y / F2 F₂_y = F2 sin 45

force 3

F3 = 171 N

cos 45 = F₃ₓ / F3

sin 45 = F₃_y / F3

the total force can be found with its components

axis x direction (East-West)

Fₓ = 64.7 +61.5 cos 45 + 171 cos 45

Fₓ = 229.1 N

Y axis (direction North -Sur)

F_y = 61.5 sin 45 - 171 sin 45

F_y = - 77,428 N

the resulting force is

F = Fx i ^ + Fy j ^

F = (229.1 i⁻ 77,428 j⁾ N

we can use the Pythagorean theorem to find the module

F = √ (229.1 2 + 77,428 2)

F = 241.8 N

let's use trigonometry to find the direction

tan θ = F_y / Fₓ

θ = tan⁻¹ (F_y / Fₓ)

θ = tan⁻¹ (-77,428 / 229.1)

θ = -18.7

4 0

3 years ago

2. A 2kg body attached to a spring undergoes a SHM of amplitude 0.4m and period

Rudik [331]

Answer:

a) 12.8 N

b) 3.2 m/s²

Explanation:

I'm guessing the period is 0.5π s.

Period of a spring in simple harmonic motion is:

T = 2π √(m/k)

Given T = 0.5π and m = 2 kg:

0.5π = 2π √(2/k)

0.25 = √(2/k)

0.0625 = 2/k

k = 32

The spring constant is 32 N/m, and the maximum displacement is 0.4 m. The maximum force can be found with Hooke's law:

F = kx

F = (32 N/m) (0.4 m)

F = 12.8 N

The acceleration can be found with Newton's second law:

∑F = ma

kx = ma

(32 N/m) (0.2 m) = (2 kg) a

a = 3.2 m/s²

8 0

3 years ago

a water line starts the service with an altitude of 1200m over the sea level, what is the velocity of the water above 1050 m ove

Ilia_Sergeevich [38]

Answer:

Velocity = 94.85m/s

Explanation:

<u>Given the following data ;</u>

Height = 1200m

Vertical distance = 1050m

To find the time, we would use the second equation of motion;

$S = ut + \frac {1}{2}at^{2}$

Substituting into the equation, we have;

$1200 = 0(t) + \frac {1}{2}*9.8*t^{2}$

$1200 = 0 + 4.9*t^{2}$

$1200 = 4.9*t^{2}$

$t^{2} = \frac {1200}{4.9}$

$t = \sqrt{122.45}$

t = 11.07 secs

To find the velocity;

Mathematically, velocity is given by the equation;

$Velocity = \frac{distance}{time}$

Substituting into the above equation;

$Velocity = \frac{1050}{11.07}$

Velocity = 94.85m/s

Therefore, the velocity of the water above 1050 m over the sea level is 94.85m/s.

7 0

3 years ago

Which source of power can reduce emission of carbon dioxide?

ki77a [65]

Answer:

wind

Explanation:

this is because the others when burnt releases large amounts of carbon of carbon dioxide as they are carbon based and oxidises when burnt. However wind is renewable and converts the kinetic energy in the wind to electrical energy by the use of a turbine

6 0

3 years ago